1 Answer
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$x=10$ is too far away to expect good results out of a series expansion of $\sqrt{1+x}$ about $x=0$. But you could take an expansion of $\sqrt{9+x}$ about $x=0$; then you know the first term of the expansion so you can continue.

Also, there is a way to reuse your work by writing $\sqrt{9+x}=a\sqrt{1+bx}$ for appropriate constants $a,b$.

$\begingroup$Great just figured out the same thing.$\endgroup$
– kembNov 7 '17 at 23:54

$\begingroup$Would it even be possible to use sqrt(11+x)?$\endgroup$
– kembNov 7 '17 at 23:55

$\begingroup$@kemb Strictly speaking yes but all that that gives you is the tautology $\sqrt{11}=\sqrt{11}$. This does not help you to evaluate the square root. You want to expand around a point where you know the answer. The two nearby points where you know the answer are $9$ and $16$; either of these will do the job (though you will get different errors).$\endgroup$
– IanNov 7 '17 at 23:57